For a finite group, there are finitely many irreducible representations of complex numbers.
What if the field is changed to some other fields? Like real numbers, p-adic field, finite field?

In particular, for a (finite) Galois group over a p-adic field, and consider p-adic Galois representation. Are there finitely many irreducible representations? If there are, can we actually construct some kind of varieteis s.t. the geometric representations (etale cohomology) coming from these varieties are exactly the irreducible ones?

And what if we replace the finite groups to other groups? Say, profinite groups, or even Lie groups, algebraic groups with non-discrete topologies?

Small point: you can always embed a p-adic field into the complex numbers and deduce finiteness in this way. For infinite, profinite groups you usually require the representation to be continuous, so and the embedding of a p-adic field into the complex numbers isn't continuous; it still turns out to be a useful thing to do though in some case (e.g. Weil-Deligne representations).
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David Zureick-Brown♦Jan 1 '10 at 20:31

thank you! but how does one deduce finiteness after tensoring with complex numbers? I mean, irreducible representations in Q_p can become reducible after tensoring with C, right? And what do you mean by working on WD representations? What are the results? Thank you!
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naturaJan 2 '10 at 2:57

There is no need to do any of these tensor product tricks. For any field $k$ of characteristic $0$, the irreducible representations of a finite group $G$ are precisely the simple submodules of the semisimple group algebra $k[G]$, so are finite in number. Some of these irreducible representations may become reducible after extension (keyword: Schur index), but that's orthogonal to the question at hand.
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Pete L. ClarkJan 3 '10 at 11:50

2 Answers
2

In characteristic zero, the group algebra is semisimple, so there are finitely many simple representations. These representations correspond to the blocks in the decomposition as a product of matrix algebras.

Here is a way to determine the number:
Start with the field $K$, adjoin the $g$th roots of unity ($g = |G|$) to get $L$, and consider the Galois group $\Gamma_K=L/K$. This is a subgroup of the multiplicative group of the integers mod $g$. Then $\sigma_t \in \Gamma_K$ corresponding to $t \in (\mathbb{Z}/g\mathbb{Z})^*$ acts on $G$ by raising $x \in G$ to the $t$-th power. The dimension of the space of class functions constant on $\Gamma_K$-orbits is the number of simple $K$-representations.

As for characteristic p, this is modular representation theory, The number of irreducibles is the number of $p$-regular conjugacy classes (where $p$-regular means the period is prime to $p$), when the field contains the $g$th roots of unity for $g$ the order of the group. See, e.g., Serre's Linear Representations of FInite Groups. My guess is that it should be true even without the assumption on the field being sufficiently large.

If a group is finite, it has finitely many simple representations over any field. Indeed, in this case the group algebra over the field is artinian, and this is true for all artinian algebras.

When the field characteristic divides the order of the group, the so called "modular situation", these finitely many representations are very far away from the whole story. For example, if the characteristic is $p$ and the group is a $p$-group, there is exactly one simple module, yet as long as the group is not cyclic, the category of representations is wild so there are extraordinarily many other indecomposable representations.